f

```Dear Torsten,

thank you very much for your kind help. i have written the code and its almost done but now only one small problem remains, here is the code that i wrote down;

function maxwellbloch
t=linspace(0,5,10);%the non dimensionalized characteritic time
z=linspace(0,5,10);%the non dimensionalized characteristic length
s=[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E(z,0) included for ten grid points, E(2),E(3),...,E(10)
[t,y]=ode45(@salman3,t,s);
meshgrid(t,z);
surf(t,z,E)
function dydt=salman3(t,y,E)
z=linspace(0,5,10);
dydt=zeros(size(y));

y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(13);y(14);y(15);y(16);y(17);y(18);
f=10; %Ten discretization points along the distance grid
E(1)=0;%Boundary condition for the wave E(0,t)=2 supposed here;
for j=10:9+f-1
E(j-10+2)=y(j);
end
for j=2:length(z)
dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
end

%--------------------------------------------------
k5=1000000000;
k2=10*i*k5/2;
k1=k2/3;
k3=100;
k4=k5/2;
k6=k4;
%------------------------------------------------------
dydt(1)=-k1*y(3)+k1*y(7)+k5*y(9);
dydt(5)=-k2*y(6)+k2*y(8);
dydt(9)=k1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
dydt(2)=-k3*y(2)-k2*y(3)+k1*y(8);
dydt(3)=-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
dydt(4)=-k3*y(4)-k1*y(6)+k2*y(7);
dydt(6)=-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
dydt(7)=k1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
dydt(8)=k1*y(2)+k2*y(5)-k6*y(8);
for j=10:9+f-1
dydt(j)=dEdt(j-10+2);
end
for m=2:length(z)
dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
end

the problem is; as you mentioned to write
for j=2:length(z)
dEdt(j)=...;
end

what should i write here? as the original equation that i wanted to solve is written in the final portion of the code, like;

for m=2:length(z)
dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
end

so i am confused with this, what should i declare here for dEdt(j)-...;

also if i want to get plot of E(z,t) vs t and z, is the meshgrid and the surf command ok that i wrote above just after calling the ode45 solver?

i will highly appreciate your help.

thanks alot.
```
 0
5/11/2011 6:12:05 AM
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```On 11 Mai, 08:12, "salman " <salmanabdull...@gmail.com> wrote:
> Dear Torsten,
>
> thank you very much for your kind help. i have written the code and its a=
lmost done but now only one small problem remains, here is the code that i =
wrote down;
>
> function maxwellbloch
> t=3Dlinspace(0,5,10);%the non dimensionalized characteritic time
> z=3Dlinspace(0,5,10);%the non dimensionalized characteristic length
> s=3D[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E(z,0) in=
cluded for ten grid points, E(2),E(3),...,E(10)
> [t,y]=3Dode45(@salman3,t,s);
> meshgrid(t,z);
> surf(t,z,E)

How should MATLAB know here what E is ?
After the call to ode45, E is hidden in y(10)-y(10+nz-1).
So write them in a two-dimensional array E where the first dimension
refers to the time step and the second dimension to the z-position.
Don't forget E(1) - it's not part of the y-vector.

> function dydt=3Dsalman3(t,y,E)

This is not the usual parameter list for the function called by ode45.
Use
function dydt=3Dsalman3(t,y)

> z=3Dlinspace(0,5,10);
> dydt=3Dzeros(size(y));
>
> y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(13);y(14=

> f=3D10; %Ten discretization points along the distance grid
> E(1)=3D0;%Boundary condition for the wave E(0,t)=3D2 supposed here;
> for j=3D10:9+f-1
> =A0 =A0 E(j-10+2)=3Dy(j);
> end
> for j=3D2:length(z)
> =A0 =A0dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
> =A0 end
>

You don't have defined i until now - and don't use i as a variable
since it usually
refers to the imaginary unit in MATLAB (or do you mean the imaginary
unit ???)

k=3D...;
c=3D...;
for j=3D2:length(z)
dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+k/c*y(3);
end

> %--------------------------------------------------
> k5=3D1000000000;
> k2=3D10*i*k5/2;
> k1=3Dk2/3;
> k3=3D100;
> k4=3Dk5/2;
> k6=3Dk4;
> %------------------------------------------------------
> dydt(1)=3D-k1*y(3)+k1*y(7)+k5*y(9);
> dydt(5)=3D-k2*y(6)+k2*y(8);
> dydt(9)=3Dk1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
> dydt(2)=3D-k3*y(2)-k2*y(3)+k1*y(8);
> dydt(3)=3D-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
> dydt(4)=3D-k3*y(4)-k1*y(6)+k2*y(7); =A0
> dydt(6)=3D-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
> dydt(7)=3Dk1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
> dydt(8)=3Dk1*y(2)+k2*y(5)-k6*y(8);
> for j=3D10:9+f-1
> =A0 =A0 dydt(j)=3DdEdt(j-10+2);
> end
> for m=3D2:length(z)
> =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> end
>
> =A0 =A0the problem is; as you mentioned to write
> for j=3D2:length(z)
> =A0 =A0dEdt(j)=3D...;
> =A0 end
>
> what should i write here? as the original equation that i wanted to solve=
is written in the final portion of the code, like;
>
> for m=3D2:length(z)
> =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> end
>
> so i am confused with this, what should i declare here for dEdt(j)-...;
>
> also if i want to get plot of E(z,t) vs t and z, is the meshgrid and the =
surf command ok that i wrote above just after calling the ode45 solver?
>

I don't have the time to read the MATLAB documentation for the graphic
commands available.
But if you have stored E in a two-dimensional array like I suggested
above and you have the
times and z-positions in two seperate one-dimensional arrays, it
should be possible to combine
them in a way to get a reasonable representation of the results.

> i will highly appreciate your help.
>
> thanks alot.

Best wishes
Torsten.
```
 0
5/11/2011 7:00:52 AM
```Torsten <Torsten.Hennig@umsicht.fraunhofer.de> wrote in message <800f8ef5-9c8e-4071-8a48-b20db4d396aa@w21g2000yqm.googlegroups.com>...
> On 11 Mai, 08:12, "salman " <salmanabdull...@gmail.com> wrote:
> > Dear Torsten,
> >
> > thank you very much for your kind help. i have written the code and its almost done but now only one small problem remains, here is the code that i wrote down;
> >
> > function maxwellbloch
> > t=linspace(0,5,10);%the non dimensionalized characteritic time
> > z=linspace(0,5,10);%the non dimensionalized characteristic length
> > s=[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E(z,0) included for ten grid points, E(2),E(3),...,E(10)
> > [t,y]=ode45(@salman3,t,s);
> > meshgrid(t,z);
> > surf(t,z,E)
>
> How should MATLAB know here what E is ?
> After the call to ode45, E is hidden in y(10)-y(10+nz-1).
> So write them in a two-dimensional array E where the first dimension
> refers to the time step and the second dimension to the z-position.
> Don't forget E(1) - it's not part of the y-vector.
>
> > function dydt=salman3(t,y,E)
>
> This is not the usual parameter list for the function called by ode45.
> Use
> function dydt=salman3(t,y)
>
> > z=linspace(0,5,10);
> > dydt=zeros(size(y));
> >
> > y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(13);y(14);­y(15);y(16);y(17);y(18);
>
> > f=10; %Ten discretization points along the distance grid
> > E(1)=0;%Boundary condition for the wave E(0,t)=2 supposed here;
> > for j=10:9+f-1
> >     E(j-10+2)=y(j);
> > end
> > for j=2:length(z)
> >    dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
> >   end
> >
>
> You don't have defined i until now - and don't use i as a variable
> since it usually
> refers to the imaginary unit in MATLAB (or do you mean the imaginary
> unit ???)
>
> k=...;
> c=...;
>  for j=2:length(z)
>     dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+k/c*y(3);
>  end
>
>
> > %--------------------------------------------------
> > k5=1000000000;
> > k2=10*i*k5/2;
> > k1=k2/3;
> > k3=100;
> > k4=k5/2;
> > k6=k4;
> > %------------------------------------------------------
> > dydt(1)=-k1*y(3)+k1*y(7)+k5*y(9);
> > dydt(5)=-k2*y(6)+k2*y(8);
> > dydt(9)=k1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
> > dydt(2)=-k3*y(2)-k2*y(3)+k1*y(8);
> > dydt(3)=-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
> > dydt(4)=-k3*y(4)-k1*y(6)+k2*y(7);
> > dydt(6)=-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
> > dydt(7)=k1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
> > dydt(8)=k1*y(2)+k2*y(5)-k6*y(8);
> > for j=10:9+f-1
> >     dydt(j)=dEdt(j-10+2);
> > end
> > for m=2:length(z)
> >     dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > end
> >
> >    the problem is; as you mentioned to write
> > for j=2:length(z)
> >    dEdt(j)=...;
> >   end
> >
> > what should i write here? as the original equation that i wanted to solve is written in the final portion of the code, like;
> >
> > for m=2:length(z)
> >     dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > end
> >
> > so i am confused with this, what should i declare here for dEdt(j)-...;
> >
> > also if i want to get plot of E(z,t) vs t and z, is the meshgrid and the surf command ok that i wrote above just after calling the ode45 solver?
> >
>
> I don't have the time to read the MATLAB documentation for the graphic
> commands available.
> But if you have stored E in a two-dimensional array like I suggested
> above and you have the
> times and z-positions in two seperate one-dimensional arrays, it
> should be possible to combine
> them in a way to get a reasonable representation of the results.
>
> > i will highly appreciate your help.
> >
> > thanks alot.
>
> Best wishes
> Torsten.

Yes exactly i is the imaginary unit here. and i understaood almost all the algo that you provided, but that one step is bugging me, i can not understand it, i will be highly grateful if y ou just clarify it a little more how should i declare E here;

for j=2:length(z)
> >    dEdt(j)=...;
> >   end, or what actually this step means?
i will do the surf and plot easily but this part is difficult for me to understand. your explanation could help me overcome my problem.

Thank you very much Torsten,

```
 0
5/11/2011 7:24:02 AM
```On 11 Mai, 09:24, "salman " <salmanabdull...@gmail.com> wrote:
> Torsten <Torsten.Hen...@umsicht.fraunhofer.de> wrote in message <800f8ef5=
> > On 11 Mai, 08:12, "salman " <salmanabdull...@gmail.com> wrote:
> > > Dear Torsten,
>
> > > thank you very much for your kind help. i have written the code and i=
ts almost done but now only one small problem remains, here is the code tha=
t i wrote down;
>
> > > function maxwellbloch
> > > t=3Dlinspace(0,5,10);%the non dimensionalized characteritic time
> > > z=3Dlinspace(0,5,10);%the non dimensionalized characteristic length
> > > s=3D[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E(z,0=
) included for ten grid points, E(2),E(3),...,E(10)
> > > [t,y]=3Dode45(@salman3,t,s);
> > > meshgrid(t,z);
> > > surf(t,z,E)
>
> > How should MATLAB know here what E is ?
> > After the call to ode45, E is hidden in y(10)-y(10+nz-1).
> > So write them in a two-dimensional array E where the first dimension
> > refers to the time step and the second dimension to the z-position.
> > Don't forget E(1) - it's not part of the y-vector.
>
> > > function dydt=3Dsalman3(t,y,E)
>
> > This is not the usual parameter list for the function called by ode45.
> > Use
> > function dydt=3Dsalman3(t,y)
>
> > > z=3Dlinspace(0,5,10);
> > > dydt=3Dzeros(size(y));
>
> > > y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(13);=
>
> > > f=3D10; %Ten discretization points along the distance grid
> > > E(1)=3D0;%Boundary condition for the wave E(0,t)=3D2 supposed here;
> > > for j=3D10:9+f-1
> > > =A0 =A0 E(j-10+2)=3Dy(j);
> > > end
> > > for j=3D2:length(z)
> > > =A0 =A0dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
> > > =A0 end
>
> > You don't have defined i until now - and don't use i as a variable
> > since it usually
> > refers to the imaginary unit in MATLAB (or do you mean the imaginary
> > unit ???)
>
> > k=3D...;
> > c=3D...;
> > =A0for j=3D2:length(z)
> > =A0 =A0 dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+k/c*y(3);
> > =A0end
>
> > > %--------------------------------------------------
> > > k5=3D1000000000;
> > > k2=3D10*i*k5/2;
> > > k1=3Dk2/3;
> > > k3=3D100;
> > > k4=3Dk5/2;
> > > k6=3Dk4;
> > > %------------------------------------------------------
> > > dydt(1)=3D-k1*y(3)+k1*y(7)+k5*y(9);
> > > dydt(5)=3D-k2*y(6)+k2*y(8);
> > > dydt(9)=3Dk1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
> > > dydt(2)=3D-k3*y(2)-k2*y(3)+k1*y(8);
> > > dydt(3)=3D-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
> > > dydt(4)=3D-k3*y(4)-k1*y(6)+k2*y(7); =A0
> > > dydt(6)=3D-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
> > > dydt(7)=3Dk1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
> > > dydt(8)=3Dk1*y(2)+k2*y(5)-k6*y(8);
> > > for j=3D10:9+f-1
> > > =A0 =A0 dydt(j)=3DdEdt(j-10+2);
> > > end
> > > for m=3D2:length(z)
> > > =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > end
>
> > > =A0 =A0the problem is; as you mentioned to write
> > > for j=3D2:length(z)
> > > =A0 =A0dEdt(j)=3D...;
> > > =A0 end
>
> > > what should i write here? as the original equation that i wanted to s=
olve is written in the final portion of the code, like;
>
> > > for m=3D2:length(z)
> > > =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > end
>
> > > so i am confused with this, what should i declare here for dEdt(j)-..=
..;
>
> > > also if i want to get plot of E(z,t) vs t and z, is the meshgrid and =
the surf command ok that i wrote above just after calling the ode45 solver?
>
> > I don't have the time to read the MATLAB documentation for the graphic
> > commands available.
> > But if you have stored E in a two-dimensional array like I suggested
> > above and you have the
> > times and z-positions in two seperate one-dimensional arrays, it
> > should be possible to combine
> > them in a way to get a reasonable representation of the results.
>
> > > i will highly appreciate your help.
>
> > > thanks alot.
>
> > Best wishes
> > Torsten.
>
> Yes exactly i is the imaginary unit here. and i understaood almost all th=
e algo that you provided, but that one step is bugging me, i can not unders=
tand it, i will be highly grateful if y ou just clarify it a little more ho=
w should i declare E here;
>
> =A0for j=3D2:length(z)> > =A0 =A0dEdt(j)=3D...;
> > > =A0 end, or what actually this step means?
>
> i will do the surf and plot easily but this part is difficult for me to u=
nderstand. your explanation could help me overcome my problem.
>
> Thank you very much Torsten,
>
> Indebted for your help,- Zitierten Text ausblenden -
>
> - Zitierten Text anzeigen -

E is part of the y-vector. You recovered it from the y-vector in the
loop
for j=3D10:9+f-1
E(j-10+2)=3Dy(j);
end
Now you have to supply the changes in time of E for ODE45.
The PDE reads (at least according to what you wrote)
dE/dt+dE/dz=3Di*y(3).
So in discretization point j (j>=3D2), the change of E in time (dEdt(j))
is
given by
dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3).
Or did I misinterpret something ?

Best wishes
Torsten.

```
 0
5/11/2011 9:02:30 AM
```Torsten <Torsten.Hennig@umsicht.fraunhofer.de> wrote in message <17e3e245-12d1-473f-8fa8-8d94b09ba943@p13g2000yqh.googlegroups.com>...
> On 11 Mai, 09:24, "salman " <salmanabdull...@gmail.com> wrote:
> > Torsten <Torsten.Hen...@umsicht.fraunhofer.de> wrote in message <800f8ef5-9c8e-4071-8a48-b20db4d39...@w21g2000yqm.googlegroups.com>...
> > > On 11 Mai, 08:12, "salman " <salmanabdull...@gmail.com> wrote:
> > > > Dear Torsten,
> >
> > > > thank you very much for your kind help. i have written the code and its almost done but now only one small problem remains, here is the code that i wrote down;
> >
> > > > function maxwellbloch
> > > > t=linspace(0,5,10);%the non dimensionalized characteritic time
> > > > z=linspace(0,5,10);%the non dimensionalized characteristic length
> > > > s=[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E(z,0) included for ten grid points, E(2),E(3),...,E(10)
> > > > [t,y]=ode45(@salman3,t,s);
> > > > meshgrid(t,z);
> > > > surf(t,z,E)
> >
> > > How should MATLAB know here what E is ?
> > > After the call to ode45, E is hidden in y(10)-y(10+nz-1).
> > > So write them in a two-dimensional array E where the first dimension
> > > refers to the time step and the second dimension to the z-position.
> > > Don't forget E(1) - it's not part of the y-vector.
> >
> > > > function dydt=salman3(t,y,E)
> >
> > > This is not the usual parameter list for the function called by ode45.
> > > Use
> > > function dydt=salman3(t,y)
> >
> > > > z=linspace(0,5,10);
> > > > dydt=zeros(size(y));
> >
> > > > y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(13);y(14);­­y(15);y(16);y(17);y(18);
> >
> > > > f=10; %Ten discretization points along the distance grid
> > > > E(1)=0;%Boundary condition for the wave E(0,t)=2 supposed here;
> > > > for j=10:9+f-1
> > > >     E(j-10+2)=y(j);
> > > > end
> > > > for j=2:length(z)
> > > >    dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
> > > >   end
> >
> > > You don't have defined i until now - and don't use i as a variable
> > > since it usually
> > > refers to the imaginary unit in MATLAB (or do you mean the imaginary
> > > unit ???)
> >
> > > k=...;
> > > c=...;
> > >  for j=2:length(z)
> > >     dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+k/c*y(3);
> > >  end
> >
> > > > %--------------------------------------------------
> > > > k5=1000000000;
> > > > k2=10*i*k5/2;
> > > > k1=k2/3;
> > > > k3=100;
> > > > k4=k5/2;
> > > > k6=k4;
> > > > %------------------------------------------------------
> > > > dydt(1)=-k1*y(3)+k1*y(7)+k5*y(9);
> > > > dydt(5)=-k2*y(6)+k2*y(8);
> > > > dydt(9)=k1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
> > > > dydt(2)=-k3*y(2)-k2*y(3)+k1*y(8);
> > > > dydt(3)=-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
> > > > dydt(4)=-k3*y(4)-k1*y(6)+k2*y(7);
> > > > dydt(6)=-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
> > > > dydt(7)=k1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
> > > > dydt(8)=k1*y(2)+k2*y(5)-k6*y(8);
> > > > for j=10:9+f-1
> > > >     dydt(j)=dEdt(j-10+2);
> > > > end
> > > > for m=2:length(z)
> > > >     dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > > end
> >
> > > >    the problem is; as you mentioned to write
> > > > for j=2:length(z)
> > > >    dEdt(j)=...;
> > > >   end
> >
> > > > what should i write here? as the original equation that i wanted to solve is written in the final portion of the code, like;
> >
> > > > for m=2:length(z)
> > > >     dEdt(m)=-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > > end
> >
> > > > so i am confused with this, what should i declare here for dEdt(j)-...;
> >
> > > > also if i want to get plot of E(z,t) vs t and z, is the meshgrid and the surf command ok that i wrote above just after calling the ode45 solver?
> >
> > > I don't have the time to read the MATLAB documentation for the graphic
> > > commands available.
> > > But if you have stored E in a two-dimensional array like I suggested
> > > above and you have the
> > > times and z-positions in two seperate one-dimensional arrays, it
> > > should be possible to combine
> > > them in a way to get a reasonable representation of the results.
> >
> > > > i will highly appreciate your help.
> >
> > > > thanks alot.
> >
> > > Best wishes
> > > Torsten.
> >
> > Yes exactly i is the imaginary unit here. and i understaood almost all the algo that you provided, but that one step is bugging me, i can not understand it, i will be highly grateful if y ou just clarify it a little more how should i declare E here;
> >
> >  for j=2:length(z)> >    dEdt(j)=...;
> > > >   end, or what actually this step means?
> >
> > i will do the surf and plot easily but this part is difficult for me to understand. your explanation could help me overcome my problem.
> >
> > Thank you very much Torsten,
> >
> > Indebted for your help,- Zitierten Text ausblenden -
> >
> > - Zitierten Text anzeigen -
>
> I don't understand your question.
> E is part of the y-vector. You recovered it from the y-vector in the
> loop
>  for j=10:9+f-1
>      E(j-10+2)=y(j);
>  end
> Now you have to supply the changes in time of E for ODE45.
> The PDE reads (at least according to what you wrote)
> dE/dt+dE/dz=i*y(3).
> So in discretization point j (j>=2), the change of E in time (dEdt(j))
> is
> given by
> dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3).
> Or did I misinterpret something ?
>
> Best wishes
> Torsten.

Yes Dear torsten,

you got it exactly right, y(1) through y(9) are initial components of the y vector. and by this statement
for j=10:9+f-1
>      E(j-10+2)=y(j); y(10) through y(nz) become the appended entries which are actually E(2),....,E(nz). and now the whole system of n+9 equations n for E(1),...,E(nz) and other for y(1),...,Y(9) are solved. also the main equation is declared in the end by the following statement;

dEdt(j)=-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);

but the only thing now thats giving bug is the following statement as you instructed me earlier is;

for j=2:length(z)
> > > >    dEdt(j)=...;
> > > >   end

i dont underdtand what should i declare here? initial conditions or boundary condiotions or what? i undertand almost all of your algorithm excpet this one point. please help me undersdtand it.

i am so much grateful for your kind help.

Thank you very much
```
 0
5/11/2011 9:31:04 AM
```On 11 Mai, 11:31, "salman " <salmanabdull...@gmail.com> wrote:
> Torsten <Torsten.Hen...@umsicht.fraunhofer.de> wrote in message <17e3e245=
> > On 11 Mai, 09:24, "salman " <salmanabdull...@gmail.com> wrote:
> > > Torsten <Torsten.Hen...@umsicht.fraunhofer.de> wrote in message <800f=
> > > > On 11 Mai, 08:12, "salman " <salmanabdull...@gmail.com> wrote:
> > > > > Dear Torsten,
>
> > > > > thank you very much for your kind help. i have written the code a=
nd its almost done but now only one small problem remains, here is the code=
that i wrote down;
>
> > > > > function maxwellbloch
> > > > > t=3Dlinspace(0,5,10);%the non dimensionalized characteritic time
> > > > > z=3Dlinspace(0,5,10);%the non dimensionalized characteristic leng=
th
> > > > > s=3D[1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1]; %The intitial values E=
(z,0) included for ten grid points, E(2),E(3),...,E(10)
> > > > > [t,y]=3Dode45(@salman3,t,s);
> > > > > meshgrid(t,z);
> > > > > surf(t,z,E)
>
> > > > How should MATLAB know here what E is ?
> > > > After the call to ode45, E is hidden in y(10)-y(10+nz-1).
> > > > So write them in a two-dimensional array E where the first dimensio=
n
> > > > refers to the time step and the second dimension to the z-position.
> > > > Don't forget E(1) - it's not part of the y-vector.
>
> > > > > function dydt=3Dsalman3(t,y,E)
>
> > > > This is not the usual parameter list for the function called by ode=
45.
> > > > Use
> > > > function dydt=3Dsalman3(t,y)
>
> > > > > z=3Dlinspace(0,5,10);
> > > > > dydt=3Dzeros(size(y));
>
> > > > > y(1);y(2);y(3);y(4);y(5);y(6);y(7);y(8);y(9);y(10);y(11);y(12);y(=
>
> > > > > f=3D10; %Ten discretization points along the distance grid
> > > > > E(1)=3D0;%Boundary condition for the wave E(0,t)=3D2 supposed her=
e;
> > > > > for j=3D10:9+f-1
> > > > > =A0 =A0 E(j-10+2)=3Dy(j);
> > > > > end
> > > > > for j=3D2:length(z)
> > > > > =A0 =A0dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
> > > > > =A0 end
>
> > > > You don't have defined i until now - and don't use i as a variable
> > > > since it usually
> > > > refers to the imaginary unit in MATLAB (or do you mean the imaginar=
y
> > > > unit ???)
>
> > > > k=3D...;
> > > > c=3D...;
> > > > =A0for j=3D2:length(z)
> > > > =A0 =A0 dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+k/c*y(3);
> > > > =A0end
>
> > > > > %--------------------------------------------------
> > > > > k5=3D1000000000;
> > > > > k2=3D10*i*k5/2;
> > > > > k1=3Dk2/3;
> > > > > k3=3D100;
> > > > > k4=3Dk5/2;
> > > > > k6=3Dk4;
> > > > > %------------------------------------------------------
> > > > > dydt(1)=3D-k1*y(3)+k1*y(7)+k5*y(9);
> > > > > dydt(5)=3D-k2*y(6)+k2*y(8);
> > > > > dydt(9)=3Dk1*y(3)+k2*y(6)-k1*y(7)-k2*y(8)-k5*y(9);
> > > > > dydt(2)=3D-k3*y(2)-k2*y(3)+k1*y(8);
> > > > > dydt(3)=3D-k1*y(1)-k2*y(2)-k4*y(3)+k1*y(9);
> > > > > dydt(4)=3D-k3*y(4)-k1*y(6)+k2*y(7); =A0
> > > > > dydt(6)=3D-k1*y(4)-k2*y(5)+k2*y(9)-k6*y(6);
> > > > > dydt(7)=3Dk1*y(1)+k2*y(4)-k4*y(7)-k1*y(9);
> > > > > dydt(8)=3Dk1*y(2)+k2*y(5)-k6*y(8);
> > > > > for j=3D10:9+f-1
> > > > > =A0 =A0 dydt(j)=3DdEdt(j-10+2);
> > > > > end
> > > > > for m=3D2:length(z)
> > > > > =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > > > end
>
> > > > > =A0 =A0the problem is; as you mentioned to write
> > > > > for j=3D2:length(z)
> > > > > =A0 =A0dEdt(j)=3D...;
> > > > > =A0 end
>
> > > > > what should i write here? as the original equation that i wanted =
to solve is written in the final portion of the code, like;
>
> > > > > for m=3D2:length(z)
> > > > > =A0 =A0 dEdt(m)=3D-(E(m)-E(m-1))/(z(m)-z(m-1))+i*y(3);
> > > > > end
>
> > > > > so i am confused with this, what should i declare here for dEdt(j=
)-...;
>
> > > > > also if i want to get plot of E(z,t) vs t and z, is the meshgrid =
and the surf command ok that i wrote above just after calling the ode45 sol=
ver?
>
> > > > I don't have the time to read the MATLAB documentation for the grap=
hic
> > > > commands available.
> > > > But if you have stored E in a two-dimensional array like I suggeste=
d
> > > > above and you have the
> > > > times and z-positions in two seperate one-dimensional arrays, it
> > > > should be possible to combine
> > > > them in a way to get a reasonable representation of the results.
>
> > > > > i will highly appreciate your help.
>
> > > > > thanks alot.
>
> > > > Best wishes
> > > > Torsten.
>
> > > Yes exactly i is the imaginary unit here. and i understaood almost al=
l the algo that you provided, but that one step is bugging me, i can not un=
derstand it, i will be highly grateful if y ou just clarify it a little mor=
e how should i declare E here;
>
> > > =A0for j=3D2:length(z)> > =A0 =A0dEdt(j)=3D...;
> > > > > =A0 end, or what actually this step means?
>
> > > i will do the surf and plot easily but this part is difficult for me =
to understand. your explanation could help me overcome my problem.
>
> > > Thank you very much Torsten,
>
> > > Indebted for your help,- Zitierten Text ausblenden -
>
> > > - Zitierten Text anzeigen -
>
> > I don't understand your question.
> > E is part of the y-vector. You recovered it from the y-vector in the
> > loop
> > =A0for j=3D10:9+f-1
> > =A0 =A0 =A0E(j-10+2)=3Dy(j);
> > =A0end
> > Now you have to supply the changes in time of E for ODE45.
> > The PDE reads (at least according to what you wrote)
> > dE/dt+dE/dz=3Di*y(3).
> > So in discretization point j (j>=3D2), the change of E in time (dEdt(j)=
)
> > is
> > given by
> > dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3).
> > Or did I misinterpret something ?
>
> > Best wishes
> > Torsten.
>
> Yes Dear torsten,
>
> you got it exactly right, y(1) through y(9) are initial components of the=
y vector. and by this statement
> =A0for j=3D10:9+f-1
>
> > =A0 =A0 =A0E(j-10+2)=3Dy(j); y(10) through y(nz) become the appended en=
tries which are actually E(2),....,E(nz). and now the whole system of n+9 e=
quations n for E(1),...,E(nz) and other for y(1),...,Y(9) are solved. also =
the main equation is declared in the end by the following statement;
>
> dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
>
> but the only thing now thats giving bug is the following statement as you=
instructed me earlier is;
>
> for j=3D2:length(z)
>
> > > > > =A0 =A0dEdt(j)=3D...;
> > > > > =A0 end
>

for j=3D10:9+f-1
dEdt(j)=3D-(E(j)-E(j-1))/(z(j)-z(j-1))+i*y(3);
end
for j=3D2:length(z)
dEdt(j)=3D...;
end
is superfluos.

> i dont underdtand what should i declare here? initial conditions or bound=
ary condiotions or what? i undertand almost all of your algorithm excpet th=
>
> i am so much grateful for your kind help.
>
> Thank you very much- Zitierten Text ausblenden -
>
> - Zitierten Text anzeigen -

Best wishes
Torsten.
```
 0
5/11/2011 10:29:39 AM

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Dear friends, i get an error ??? Error using ==> photonechofdtd>echo Too many output arguments. function photonechofdtd tau=0:29; zee=0:29; f=30; L=30;%Length of the medium alpha=1;% Absorption coefficient s=[.4,.6,0,0]; for h=2:f s(4+5*(h-2)+1)=1-.6*exp(-zee(h)); s(4+5*(h-2)+2)=.6*exp(-zee(h)); s(4+5*(h-2)+3)=0; s(4+5*(h-2)+4)=0; end [T,Y]=ode15s(@(t,y)echo(t,y),tau,s);<--------------Here it gives error. for w=1:numel(T) echo_i(w,1)=0; for h=2:f echo_i(w,h)=Y(w,4+5*(h-2)+2).*exp(Y(w,4+5*(h-2)+2)-Y(w,4+5*(h-2)+1)).*(L-zee(h)); end end size(echo_i) size(tau) size(g) function echo(tau,y) omega=5000000; f=0:29; zee=0:29; y1(1) = y(1); y2(1) = y(2); y3(1) = y(3); y4(1) = y(4); for h=2:f y1(h) = y(4+5*(h-2)+1); y2(h) = y(4+5*(h-2)+2); y3(h) = y(4+5*(h-2)+3); y4(h) = y(4+5*(h-2)+4); end dydt =zeros(numel(y),1); dy1dt=zeros(numel(zee),1); dy2dt=zeros(numel(zee),1); dy3dt=zeros(numel(zee),1); dy4dt=zeros(numel(zee),1); for h=1:f dy1dt(h)=-(i/2)*omega*(y3(h)-y4(h)); dy2dt(h)=(i/2)*omega*(y3(h)-y4(h)); dy3dt(h)=-(i/2)*omega*(y1(h)-y2(h)); dy4dt(h)=(i/2)*omega*(y1(h)-y2(h)); end dydt(1) = dy1dt(1); dydt(2) = dy2dt(1); dydt(3) = dy3dt(1); dydt(4) = dy4dt(1); for h = 2: f dydt(4+5*(h-2)+1) = dy1dt(h); dydt(4+5*(h-2)+2) = dy2dt(h); dydt(4+5*(h-2)+3) = dy3dt(h); dydt(4+5*(h-2)+4) = dy4dt(h); end Dear friends, the error is only in a line, but i pasted my code so that you may look easily. Thanks, Waiting for...

Dear friends, i have a program in MATLAB, that requires me to run it again and again by changing different parameters. is it possible that i may not have to run the program again and again and just write the plot command in the command window (rather than calling the m file again and again)? rephrasing my question, if in my code i have to extract various vectors from this vector Y(:,:,:), is it possible that when i call the m file once and then aftewards i store the matrix Y(:,:,:) data somehow and then giving the plot command in just the command window (without calling the m file again) and change the parameters to see the new result? i think i made it clear. i will wait for your help. Thanks alot On Sep 10, 11:40=A0am, "salman " <salmanabdull...@gmail.com> wrote: > Dear friends, > > i have a program in MATLAB, that requires me to run it again and again by= changing different parameters. is it possible that i may not have to run t= he program again and again and just write the plot command in the command w= indow (rather than calling the m file again and again)? rephrasing my quest= ion, if in my code i have to extract various vectors from this vector Y(:,:= ,:), is it possible that when i call the m file once and then aftewards i s= tore the matrix Y(:,:,:) data somehow and then giving the plot command in j= ust the command window (without calling the m file again) and change the pa= rameters to see the new result? > > i think i...

Hi everyone! I'm new to Matlab and currently trying to solve an ODE system...and something is not working because I'm having a NaN solution vector. Here's my main test.m file: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% initial values vector: y0=zeros(5,1); y0(1) = 0; y0(2) = 0; y0(3)= 33.3; y0(4)= 8.307; %Plage de temps de simulation tspan=[0,10000]; %Appel du solveur d'équations différentielles (ODE) [T,Y] = ode15s(@equation_test, tspan, y0); %I tried ode45, ode23s and ode114.. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% And the equation_test.m file containing my differential equations : function dydt=equation_test(t,y) dydt=zeros(5,1); k1=0.003;k_1=0.06; k2=0.01;k_2=0.1; k3=1;k_3=0.01; Vmax4=450;Km4=50; v1=k1*y(3)*y(4)-k_1*y(1); v2=k2*y(1)*y(1)-k_2*y(2); v3=k3*y(2)-k_3*y(5); v4=Vmax4*(y(5)/(Km4*y(5))); %%%% The problem is probably here..0/0 giving NAN %%%% but how can I fix this if y(5) has to equal to %%%% 0 for the first iteration ??? dydt(1)=v1-(2*v2); dydt(2)=v2+v3-v4; dydt(3)=-v1; dydt(4)=-v1; dydt(5)=-v1; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Thanks for your help ! "Genevieve " <genevieve_mass02@hotmail.com> wrote in message <iavg8n\$rev\$1@fred.mathworks.com>... > v4=Vmax4*(y(5)/(Km4*y(5))); %%%% The problem is probably here..0/0 giving NAN > ...

FRIENDS, here is a portion of my code. in the * line (3rd last line) there is an error in interp1 call. as i think, i have inserted every thing correctly but it gives error. i can not understand it. please help me ion this. thanks alot t=0:.00001:.0001; s=[1,0,0,0]; e=[1,0]; r=0:.00001:.0001; r1=0:.00001:.0001; k_r=r; t1=0:.00001:.0001; dd=(t(end)-t(1))/length(t); k=1*pi/(10.95*.000003); td1=.000003; td2=td1+.0000064; k2=10*pi/(.00002);% this Rabi frequency in combination with the Pulse duration gives the pi pulse area tr1=.000043; tr2=tr1+.000008; zeros1d=td1/dd; zeros1d=round(zeros1d) onesd=(td2-td1)/dd; onesd=round(onesd) zeros2d=(.0001-td2)/dd; zeros2d=round(zeros2d) zeros1r=tr1/dd; zeros1r=round(zeros1r) onesr=(tr2-tr1)/dd; onesr=ceil(onesr) zeros2r=(.0001-tr2)/dd; zeros2r=round(zeros2r) k_d=k*[zeros(1,zeros1d),.8470*ones(1,onesd),zeros(1,zeros2d)]; k_r=k2*[zeros(1,zeros1r),.25090*ones(1,onesr),zeros(1,zeros2r)];% pi sca size(k_d) size(k_r) k_c=k_d+k_r; a=340000; x=linspace(-2*a,2*a,161); del=(1/(sqrt(pi)*a))*exp(-.5*(x./a).^2); del=del/sum(del); for u=1:length(del) [T,Y]=ode23s(@(t,y)system1(t,y,x(u),t1,k_c),t,s); Y_c3(:,:,u)=Y(:,:); end save Y_c3 for d=1:length(t) final1(d)=sum(real(Y_c3(d,1,:))); end for j=1:length(t) final2(j)=sum(real(Y_c3(j,2,:))); end for k=1:length(final2) [R,A]=ode45(@(r,a)second(r,a,final2(k),r1,k_r),r,e); P(:,:,k)=a(:,:); end save P; function dydt=system1(t,y,del,t1,k_c)...

Hi, im trying to figure out how to save a matrix(image) without losing any information of the matrix. Lets me explain myself: If i have a matriz like this: f=127*ones(20,20); And then i show it imshow(f) Its a white square.....this is incorrect because if: 0 = white 255=black then 127 = grey!!!! Now if we do this: imshow(f,[0 255]) It shows the grey square because we have specified exactly the range of grayscale... THE PROBLEM IS: I can save the image because it always happen the same thing that occur in our first case, i mean, we lose the information and it will save a white square!!! I beg you fellas, if anyone knows how to save the matrix as an image(imwrite) how to do it without losing the info. THANKS "Dave" <kohonen@hotmail.com> wrote in news:2245807.1140198806730.JavaMail.jakarta@nitrogen.mathforum.org... > Hi, im trying to figure out how to save a matrix(image) without losing any information of the matrix. [...] > f=127*ones(20,20); It looks like your f image is an indexed image. The color depend of the colormap. Look at the imwrite documentation. > imshow(f) > Its a white square.....this is incorrect because if: imwrite(f,'f.bmp','bmp'); produce also a white square. [...]> imshow(f,[0 255]) > It shows the grey square because we have specified exactly the range of grayscale... > m=[0:255]'*[1 1 1]/255; % 256 length gray scale imwrite(f,m,'fm.bmp'); produce a middle gray s...

Dear Friends, Hi, i am having a problem. i have an equation; dA(z,t)/dz=-a/2)*cos(B(z))*A(z,t)+a*sin(C(z))*F*(2t1-t2-t); here B, C are function of z only, while A is function of z and t. others are constant. how may i plot this? just with using some ode solver? this function should give me the result of A as function of space and time. will some FDTD method be used here? thanks alot for help. "salman " <salmanabdullah9@gmail.com> wrote in message <iu6p5k\$9p2\$1@newscl01ah.mathworks.com>... > Dear Friends, > > Hi, i am having a problem. i have an equation; > > dA(z,t)/dz=-a/2)*cos(B(z))*A(z,t)+a*sin(C(z))*F*(2t1-t2-t); > > here B, C are function of z only, while A is function of z and t. others are constant. how may i plot this? just with using some ode solver? this function should give me the result of A as function of space and time. will some FDTD method be used here? > > thanks alot for help. - - - - - - - - - - - - You can regard this as an ordinary differential equation in z, with t being held fixed. For each possible value of t, if you have an initial condition for A(z,t) at some z, this differential equation can be solved over the desired range of z while holding t at that fixed value. I think it would be rather tedious carrying out such a process with numerous different t values using one of the 'ode' solvers, though that remains as a distinct possibility. In a sense this can also be regarded as...

Dear friends, i have a system, dE/dz+k*dE/dt=k*y(3); here y(3) is a vector. i am trying it using other methods also but one way to do it would be through the advection equation. however the standard advection equation has often a zero or a constant term on the RHS, while i have a vector quantity on the RHS. any idea or suggestion how to add the vector term to the RHS of the advection equation in each iteration? thanks alot On 5/15/2011 8:19 PM, salman wrote: > Dear friends, > > i have a system, dE/dz+k*dE/dt=k*y(3); here y(3) is a vector. i am trying it > using other methods also but one way to do it would be through the advection equation. > however the standard advection equation has often a zero or a constant term on the RHS, > while i have a vector quantity on the RHS. > > any idea or suggestion how to add the vector term to the RHS of the advection > equation in each iteration? > thanks alot It looks like you have a set of N pde's, which each can be solved on its own. Using the i'th entry in the vector y for the i'th pde. So, now you have N pde's, each has a scalar in the RHS. Not a vector. Solve them one by one, in a loop. --Nasser On May 16, 3:19=A0pm, "salman " <salmanabdull...@gmail.com> wrote: > Dear friends, > > i have a system, dE/dz+k*dE/dt=3Dk*y(3); here y(3) is a vector. i am tryi= ng it using other methods also but one way to do it would be through the ad= vection equation. h...

problem with check if number is integer in matlab? help please
Hi, I have a big problems in matlab, with checking if number is integer. Part of code from my function: cijeli=zeros(size(rjesenje,1),1); for i=1:size(rjesenje,1) if mod(rjesenje(i,1),1)==0 cijeli(i,1)=1; end end When I list vector rjesenje, it shows: 1.0000 1.0000 3.0000 2.0000 1.0000 but cijeli is (after loop): 0 1 1 1 1 It shoulkd be : 1 1 1 1 1 (in vert) What's hapenning here? When I try this: mod(rjesenje(1,1),1) it shows me 1.000 but rjesenje(1,1) is 1.000 ???? Can anybody help me? Is there some mark which matlab sets to the begin of vector, so that is problem? Thanks in advance, Amer On 5/27/2011 5:04 PM, Amer wrote: .... > I have a big problems in matlab, with checking if number is integer. > Part of code from my function: > > cijeli=zeros(size(rjesenje,1),1); > > for i=1:size(rjesenje,1) > if mod(rjesenje(i,1),1)==0 cijeli(i,1)=1; > end > end > > When I list vector rjesenje, it shows: > > 1.0000 > 1.0000 > 3.0000 > 2.0000 > 1.0000 .... try format long and then look at the result of rjesenje; I think you'll begin to get the idea of what the matter is... > Can anybody help me? This is a typical problem w/ floating point representation when a nearly integer value is obtained by some other operation than actually entering an integer. The internal representation isn't _quite_ exactly 1 in this case but very close. See the FAQ at <http://matlabwiki.mathworks.co...

Dear Friends, i need your help in a following problem. my final target is to get a profile of a quantity YY(x,t) whic is connected with 2 ode solvers in a specific way. please have a look at this short code, this will make it clear what i want to do. for m=1:length(del) [T,Y]=ode45(@(t,y)system(t,y,del(m),t1,k1,t2,k2),t,s)%k1 and k2 are time dependent quantities Y1(:,:,m)=Y(:,:) end for m=1:length(del) YY(m)=sum(real(Y1(m,1,:))) end for g=1:length(del) YYY(g)=sum(real(Y1(m,2,:))) end function system(t,y,t1,k1,t2,k2) k1=interp1(t1,k1,t); k2=interp1(t2,k2,t); dydt(1)=k1*y(1)+del*y(2); dydt(2)=k2*y(2)+del*y(1); %--------------------------------------------% now these quantites YY and YYY are used in the following differential equations wrt. x as following; dydx(1)=-exp(-x)*YY(1);% these equations have to be solved in x for every tvalue of time dydx(2)=exp(-x)*YYY(1); and i want now to solve these two equations in space for every value of time in previous coupled equations. is this approach correct? please tell me friends how can i do this? thanks "salman " <salmanabdullah9@gmail.com> wrote in message <jp73ci\$5t2\$1@newscl01ah.mathworks.com>... > Dear Friends, > > i need your help in a following problem. my final target is to get a profile of a quantity YY(x,t) whic is connected with 2 ode solvers in a specific way. please have a look at this short code, this will make it clear what i want to do. > > for...

Hi all, Just bought an Evesham A410 Voyager which comes with the AMD Turion MT34 cpu and the ATI mobile X700 graphics card. I would like to play the occasional game of Counter Strike Source when I'm working away but have found it jumpy when I've played. I've checked all settings etc and the system hould be up to a game - I'm guessing its the 512mb of ram. I'm new to laptops etc and need some advice on ram - Do they use pc3200 ddr (SODimm)? Is my notebook capable of running dual channel? what ram do I buy? Thanks in advance. An excellent place to purchase ram is from Crucial. Go to their website, enter your make and model and it will tell you which and how much ram your computer will be able to use. www.crucial.com -- Don Vancouver, USA "Chris-John Turner" <sacrum@clara.co.uk> wrote in message news:1122821279.3182.0@lotis.uk.clara.net... > Hi all, > > Just bought an Evesham A410 Voyager which comes with the AMD Turion MT34 > cpu > and the ATI mobile X700 graphics card. I would like to play the > occasional > game of Counter Strike Source when I'm working away but have found it > jumpy > when I've played. I've checked all settings etc and the system hould be > up > to a game - I'm guessing its the 512mb of ram. I'm new to laptops etc and > need some advice on ram - Do they use pc3200 ddr (SODimm)? Is my notebook > capable of running dual channel? what ram...

HI, I installed php4 for apached and restart apache afterward. but my little php script generated error followint error: PHP Notice: Undefined index: myname in /usr/local/www/data-dist/www.authtec.com/php-test.php on line 2 I was trying to run the following php script in apache: <?php if(\$_POST["myname"]) { print "Hello, ".\$_POST["myname"]."<p>\n" .nl2br(\$_POST["textbox"]); } else { ?> <form method="post" action="php-test.php"> <input type=input type=text name=myname> <textarea cols=20 rows=4 wrap=virtual name=textbox></textarea> <input type=submit value="Submit"> </form> <?php } ?> What could be wrong with my configuration? and how can I fix it? Thanks Sam sam wrote: <snip> > <?php > if(\$_POST["myname"]) { > print "Hello, ".\$_POST["myname"]."<p>\n" > .nl2br(\$_POST["textbox"]); > } else { <snip> You are getting a notice, not an error. Do this instead: if(isset(\$_POST["myname"])) and your notice will be gone. Sam, You should _always_ make sure a key exists in an array before using it. <?php if (array_key_exists('myname', \$_POST) ) { echo "<p>Hello, ", \$_POST['myname'], "</p>"; } ?> I also would s...

I am a newbie to Castor. I downloaded all the jar files, made changes to the classpath etc, but I keep on getting exceptions being thrown. The XML schema file is as follows: <xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <xsd:element name="catalog"> <xsd:complexType> <xsd:sequence> <xsd:element name="book" minOccurs="0" maxOccurs="unbounded"> <xsd:complexType> <xsd:sequence> <xsd:element name="author" type="xsd:string"/> <xsd:element name="title" type="xsd:string"/> <xsd:element name="genre" type="xsd:string"/> <xsd:element name="price" type="xsd:float"/> <xsd:element name="publish_date" type="xsd:date"/> <xsd:element name="description" type="xsd:string"/> </xsd:sequence> <xsd:attribute name="id" type="xsd:string"/> </xsd:complexType> </xsd:element> </xsd:sequence> </xsd:complexType> </xsd:element> </xsd:schema> If I type in: C:\>java org.exolab.castor.builder.SourceGenerator -i C:\\Temp\\Test.xsd I get the error message: Exception in thread "main" java.lang.I...

i want help to find this Equations in matlab Quickly for my research (statistics about image) and the Answer im my mail Please
how i foud this Equation to To find inf about image and how How to apply those equations in matlab 1-mean2 avarage 2-std2 standard deviation 3-corr2 correlation 4-entropy 5-length 6-RMSE 7-SNR 8-PSN "ahmed abad" <princeofpop46@googlemail.com> wrote in message <h4n8pe\$rvv\$1@fred.mathworks.com>... > how i foud this Equation to To find inf about image and how How to apply those equations in matlab > 1-mean2 avarage > 2-std2 standard deviation > 3-corr2 correlation > 4-entropy > 5-length > 6-RMSE > 7-SNR > 8-PSN ...

Hi all, I think it might be my original database table after all that caused the problem. Unlike I said below, I pasted the columns containing the data from my original database into a new table and I kept having the problem. I just tried to actually fill the new table with fresh data and now I can run the query smoothly. Thanks for the help, I think I have to check on the integrity of the data in the original table. Maarten "Maarten van der Cammen" <maartenvandercammen@yahoo.com> wrote in message news:... > Hi Barry, > > When I switch to the SQL view of my query, this is what I have: > > SELECT TableTest.testID, [FieldA]/[FieldB] AS divided > FROM TableTest > WHERE ((([FieldA]/[FieldB])>0.5)); > > In the query view it look like: > > > > When I run the query, after two time "Overflow" message I get: > > > > > Also I started a new database, filled in some data to find out whether the > database I work with might contain some corrupt data or something, but I > kept having the same problem. > It might be that it is just not possible with double precision data in > Access to perform such actions, but to me that sounds pretty weird. > > Thanks for the help. > > Maarten > > > > > "Barry Edmund Wright" <bwright_msaccess@sympatico.ca> wrote in message > news:1181437818.310656.292570@p47g2000hsd.googlegroups.com... >> On Jun...

I want to output an arbitrary waveform using LabView onto the oscilliscope through the DAQ2005 card. Can anyone help please?
Hey guys, &nbsp; I am new to LabView and my final year project depends on it! Please help as much as you can! &nbsp; I am trying to&nbsp;output an arbitrary waveform on an oscilliscope. The shape of the waveform&nbsp;depends on user input parameters like no. of samples, amplitude of each sample and the&nbsp;frequency of desired waveform. Please refer to attached Arbitrary_Wave_Display.vi. In the attached vi, the desired waveform is displayed on the front panel. My project requires that I output this waveform through the analogue output channels of the 2 DAQ2005 cards that I have. When I connect the output of th Arbitrary_Wave.vi to the input of the AO write express vi, there is no output of the oscilliscope. I believe it should be connection problems with regards to while loops and loop conditions. I have also attached my attempted vi. &nbsp; Please feel free to give some advice. It is greatly appreciated, &nbsp; Best regards, TheApprentice of LabView &nbsp; &nbsp; ...

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